Chapter 28 – AC Circuits

LC Circuit •  Suppose switch closed at t = 0 with capacitor initially

charged to Q. What is q(t)?

VC =q

C

Ld2q

dt2+

q

C= 0

q(t) = Q cos(ωt)

ω =�

1LC

q(t = 0) = Q

I(t = 0) = 0

Initial conditions:

+q -q

I

��E · �dl = −L

dI

dt= −

�

i

∆Vi = −Vc

I = −dq

dt

•  Initially, energy is stored by capacitor in the form of an electric field.

•  When capacitor is drained (q=0), what form is the energy? •  Notice that

•  What form is energy in inductor? •  Magnetic! for solenoid:

•  Magnetic energy density:

•  Always true (general result is based on a much more formal derivation)

when q=0, |I| = Imax = Qω = Q

�1

LC

U =12CV 2 =

12

Q2

C

All energy in inductor!

12LI2 = uB × volume

uB =B2

2µ0

U =1

2

Q2

C=

1

2LI2max

•  Energy in an LC circuit oscillates between electric and magnetic •  similar to mechanical harmonic oscillator (kinetic and potential)

Damped LC oscillations •  With resistance in the circuit, LC oscillations damp out.

•  underdamped case:

•  Similar to a damped mass on spring

I

��E · d� = VC + IR = −L

dI

dt

Ld2q

dt2+ R

dq

dt+

1C

q = 0

q(t) = q0 exp(−R/(2L)t) cos

��1

LC− R2

4L2t

�

+q

-q I

I =dq

dt

Driven AC Circuits •  Driven resistor:

•  Current in phase with driving voltage source.

Vemf (t)− IR = 0 I =Vemf

R

I

+ -

I =dq

dt= C

dVemf

dt= VpCω cos(ωt) = VpCω sin(ωt + π/2)

Imax =Vp

XC

=⇒ XC =1

ωCCapacitive reactance

Vemf − VC = 0

VC =q

C= Vemf

Driven Capacitor

•  A light bulb is attached in series to a capacitor and an AC voltage source with an adjustable frequency. Keeping the peak voltage constant, the light bulb will be brightest for

1.  low frequencies 2.  high frequencies 3.  same for all frequencies

C

R

(t)

3Driven RC Circuit

ξ(t)− IR− q

C= 0

ξ = Vp sin(ωt)

dq

dtR+

q

C= Vp sin(ωt)

I =dq

dt=

Vp�( 1ωC )2 +R2

cos(ωt+ δ)δ = tan−1(−ωRC)

•  Faraday’s Law of induction

Driven Inductor

��E · d� = −L

dI

dt= −Vemf

I(t) = − Vp

ωLcos(ωt)

Imax =Vp

XL

XL = ωL Inductive reactance

Driven LR Circuit

−V (t) + IR = −LdI

dt

I(t) =V0�

R2 + (Lω)2cos(ωt− φ)

φ = tan−1 wL

R

if V (t) = V0 cos(ωt)

Jumping ring demo •  why does ring always repel from solenoid (with AC

circuit)?

http://www.youtube.com/watch?v=Pl7KyVIJ1iE

Jumping Ring Analysis

•  Suppose current in coils goes as I(t) = I0 sin(ωt) •  The emf in loop will go as

•  The induced current will therefore go as

•  Repulsive force when Iring and I are in opposite directions

I(t)

φ = tan−1 wL

R

Vemf ∝ −dI/dt ∝ − cos(ωt)

Iring ∝ − cos(ωt− φ)

Jumping Ring Cont’d •  If there were no phase difference between induced

current and emf, induced current would be in same direction as I(t) half the time, and the ring would not jump!

I(t)

Vemf

Jumping Ring Cont’d •  Now add phase shift between induced emf and induced

curent, given by

•  For large inductance/and or high frequencies, phase shift is near 90°. Here is a phase shift of 70°:

φ = tan−1 wL

R

I(t) Vemf

Iind

Now currents are in opposite directions more often than in same direction!

speaker crossover

Driven LRC Circuit

•  An RLC circuit driven by an AC voltage source exhibits frequency-dependent behavior.

−V (t) + IR +q

C= −L

dI

dt

Ip =Vp�

R2 + (XL −XC)2=

Vp

Z

I =dq

dt

+q -q

Ld2q

dt2+R

dq

dt+

q

C= Vp sin(ωdt)

Ip =Vp�

R2 + (ωL− 1ωC )2

I(t) = Ip sin(ωdt− φ)

tanφ =XL −XC

R

V (t) = Vp sin(ωdt)

Amplitude and phase in the driven RLC circuit

•  Peak current has a maximum at resonance: •  resonant frequency

•  Ip = Vp/R •  At resonance, voltage and

current are in phase

(XL = XC =�

L

C)

Transformers and power supplies •  A transformer is a pair of

coils linked by mutual inductance.

•  An AC current in the primary induces a current in the secondary.

•  The secondary voltage differs from the primary voltage by the ratio of the number of turns.

•  Both step-up and step-down transformers are possible.

ξ1 = −N1dΦB

dt(ΦB is flux per winding)

ξ2 = −N2dΦB

dt

ξ2

ξ1=

N2

N1=

V2

V1

|V1|max = |ξ1|max

|V2|max = |V1|maxN2

N1

Question 34.18 Transformers I

A 6 V battery is connected to one

side of a transformer. The output

voltage across coil B is:

Question 34.20 Transformers III

A B 6 V

•  Transformers are used to produce low voltages for electronic equipment.

•  Then they’re combined with diodes that convert AC to DC and capacitors to smooth the DC voltage.

Input Power Vs Output Power •  How does the input power compare to the output power?

•  Ignoring self inductance, P1 = V1I1

P1 = P2

I2 = I1V1

V2= I1

N1

N2

Clicker Question If the resistance R is decreased what happens to I1?

a) Increases b) Decreases c) Remains the same

Transmission of Electrical Power

Slide 26-11

•  Electric power is most efficiently transmitted at high voltages. •  This reduces I2R energy losses in the power lines. •  But most end uses require lower voltages. •  Transformers accomplish voltage changes throughout the power grid.

Two-Phase Power to Your Home

Slide 26-12

center tap is connected to ground

rms = root mean squared: